Question

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by `1/15`
. Find the fraction

Answer 1

Let the denominator of the required fraction be x.
Numerator of the required fraction =x-3 

∴Original fraction =`(x-3)/x` 

If 1 is added to the denominator, then the new fraction obtained is `(x-3)/(x+1)`  

According to the given condition, 

`(x-3)/(x+1)=(x-3)/x-1/15` 

⇒`(x-3)/x-(x-3)/(x+1)=1/15` 

⇒`((x-3)(x+1)-x(x-3))/(x(x+1))=1/15` 

⇒`(x^2-2x-3-x^2+3x)/(x^2+x)=1/15`

⇒`(x-3)/(x^2+x)=1/15` 

⇒`x^2+x=15x-45` 

⇒`x^2-14x+45=0` 

⇒`x^2-9x-5x+45=0` 

⇒`x(x-9)-5(x-9)=0` 

⇒`(x-5)(x-9)=0` 

⇒`x-5=0  or  x-9=0` 

⇒`x=5  or  x=9` 

When `x=5` 

`(x-3)/x=(5-3)/5=2/5` 

When `x=9`  

`(x-3)/x=(9-3)/9=6/9=2/3`  (This fraction is neglected because this does not satisfies the given  condition.) 

Hence, the required fraction is `2/5`